(a)(i) Coulomb's law of electrostatics The electrostatic force between two charged bodies is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. (ii) Calculation of the magnitude of the electrostatic force F = \(\frac{kq_{1}q_{2}}{d^{2}}\) F = \(\frac{9.0 \times 10^{9} \times 1.6 \times 10^{-19} \times 1.6 \times 10^{-19}}{(5.2 \times 10^{-11})^{2}}\) = 8.52 x 10\(^{-8}\) N (b)(I) Proof of V\(_{out}\) Let the current through the circuit be I V\(_{in}\) = I(\(R_1\) + \(R_2\)) OR I = \(\frac{V_{in}}{R_{1} + R_{2}}\) V\(_{out}\) = I\(R_2\) II. Calculation of \(R_1\); V\(_{out}\) = V\(_{in}\)(\(\frac{R_{1}}{R_{1} + R_{2}}\)) OR \(\frac{V_{out}}{V_{in}} = \frac{R_{1}}{R_{1} + R_{2}}\) \(\frac{1}{2.5} = \frac{R_{1}}{R_{1} + R_{30}}\) \(R_1\) = 20\(\Omega\) (ii) The Volt The unit of potential difference between two points when one joule of work is done in taking one coulomb of charge between the points, (c) Why wood is not suitable for use as the Core of transformers Wood is a non-magnetic material and would not provide magnetic coupling between the coils for electromagnetic induction. (d) Applications of cathode ray tube i. For studying alternating current waveform ii. For measuring frequency/voltage/ reactance/power factor and resistance of an electronic device iii. Used to monitor signals/check FM/AM characteristics